Algorithmic Variations on Linear Differential Equations

Bruno Salvy Inria & ENS de Lyon

MBM’2014

Computer Algebra

• Systems with millions of users

• A scientific area: effective mathematics and their complexity

• 30 years of progress in mathematical algorithms

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Thesis in this talk: linear differential equations are a good data-structure.

Menu dégustation

1. Equations as a data-structure

2. Guess & Prove combinatorial walks

3. Proofs of identities

4. Sums and Integrals

5. Numerical evaluation via the Taylor series

6. Chebyshev expansions

I. Equations as adata-structure

erf := (y00 + 2xy

0 = 0, ini. cond.)

basis of the gfun package

Dynamic Dictionary of Mathematical Functions

• User need

• Recent algorithmic progress

• Maths on the web

http://ddmf.msr-inria.inria.fr/

Heavy work by F. Chyzak

Demonstration

II. Guess & Prove Combinatorial Walks

Gessel's walks in the 1/4-plane

G(x, y, t) :=X

n�0

X

i,j

fi,j;nxiy

jt

n

• 79 inequivalent step sets (MBM & Mishna); • long history of special cases; • Gessel’s was left; • conjectured not soln of LDE.

Thm. [Bostan-Kauers 2010]G is algebraic!

Computer-driven discovery & proof.

Computation

G(x, y, t) :=X

n�0

X

i,j

fi,j;nxiy

jt

n

• Compute G up to t1000; • conjecture LDE with 1.5 million coeffs! • check for sanity (bit size, more coeffs, Fuchsian, p-curvature); • oh! • conjecture polynomials (deg ≤ (45,25,25), 25 digit coeffs); • Proof by (big) resultants.

The 79 cases: finite and infinite groups

79 step sets

23 admit a finite group[Mishna’07]

56 have an infinite group[Bousquet-Melou-Mishna’10]

all are holonomic19 transcendental[Gessel-Zeilberger’92]

[Bousquet-Melou’02]

4 are algebraic(3 Kreweras-type + Gessel)

[BMM’10] + [B.-Kauers’10]

�! all non-holonomic• [Mishna-Rechnitzer’07] and

[Melczer-Mishna’13] for 5 singular cases

• [Kurkova-Raschel’13] and

[B.-Raschel-Salvy’13] for all others

20/45

[Slide borrowed from A. Bostan]

Proving no LDE exists

• log(n), na (a∉Z), pn, e√n, Γ(n√2),... by their asymptotic behavior, or that of their generating function (with P. Flajolet & S. Gerhold).

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• walks in the 1/4-plane (with A. Bostan & K. Raschel) � :=

X

(i,j)2S

x

iy

j

c :=@2�@x@yq

@2�@x2 · @2�

@y2

(x0

, y0

)

arccos(c)/⇡ 62 Q

III. Proofs of Identities

Proof technique> series(sin(x)^2+cos(x)^2-1,x,4);

O(x4)

Why is this a proof?

1. sin and cos satisfy a 2nd order LDE: y’’+y=0; 2. their squares and their sum satisfy a 3rd order LDE; 3. the constant -1 satisfies y’=0; 4. thus sin2+cos2-1 satisfies a LDE of order at most 4; 5. Cauchy’s theorem concludes.

Proofs of non-linear identities by linear algebra!

f satisfies a LDE⟺

f,f ’,f ’’,… live in a finite-dim. vector space

Example: Mehler’s identity for Hermite polynomials

1X

n=0

Hn

(x)Hn

(y)un

n!=

exp

⇣4u(xy�u(x2+y

2))1�4u

2

⌘

p1� 4u2

1. Definition of Hermite polynomials: recurrence of order 2;

2. Product by linear algebra: Hn+k(x)Hn+k(y)/(n+k)!, k∈ℕgenerated over (x,n) by → recurrence of order at most 4;

3. Translate into differential equation.

QHn(x)Hn(y)

n!,Hn+1(x)Hn(y)

n!,Hn(x)Hn+1(y)

n!,Hn+1(x)Hn+1(y)

n!

Guess & prove continued fractions (with S. Maulat)

arctan x =x

1+13x

2

1+415x

2

1+935x

2

1+ · · ·

1. Differential equation produces first terms (easy):

2. Guess a formula (easy): an =n2

4n2 � 1

3. Prove that the CF with these an satisfies the differential equation.

No human intervention needed.

Taylor Continuedfraction

Automatic Proof of the guessed CF

• Aim: RHS satisfies (x2+1)y’-1=0; • Convergents Pn/Qn where Pn and Qn satisfy a LRE

(and Qn(0)≠0); • Define Hn:=(Qn)2((x2+1)(Pn/Qn)’-1); • Hn is a polynomial in Pn,Qn and their derivatives; • therefore, it satisfies a LRE that can be computed; • from it, Hn=O(xn) visible, ie lim Pn/Qn soln; • conclude Pn/Qn➝ arctan (check initial cond.).

arctan x?=

x

1+· · ·

1+n2

4n2�1x2

1+ · · ·

More generally: this guess-and-proof approach applies to CF for solutions of (q-)Ricatti equations → all explicit C-fractions in Cuyt et alii.

IV. Sums and Integrals

Creative telescoping

Input: equations (differential for f or recurrence for u).

Output: equations for the sum or the integral.

Method: integration (summation) by parts and differentiation (difference) under the integral (sum) sign

Example (with Pascal’s triangle):

Algorithms: reduce the search space.

I(x) =

Zf(x, t) dt =? or S(n) =

X

k

u(n, k) =?

u(n, k) =

✓n

k

◆def. by

⇢✓n+ 1

k

◆=

n+ 1

n+ 1� k

✓n

k

◆,

✓n

k+ 1

◆=

n� k

k+ 1

✓n

k

◆�

S(n+ 1) =X

k

✓n+ 1

k

◆=

X

k

✓n+ 1

k

◆�✓n+ 1

k+ 1

◆

| {z }telesc.

+

✓n

k+ 1

◆�✓n

k

◆

| {z }telesc.

+2

✓n

k

◆= 2S(n).

Telescoping Ideal

• hypergeometric summation: dim=1 + param. Gosper. [Zeilberger]

• holonomy: restrict int. by parts to and Gröbner bases. [Wilf-Zeilberger]

• finite dim, Ore algebras & GB (with F. Chyzak)

• infinite dim & GB (with F. Chyzak & M. Kauers)

• rational f and restrict to in very good complexity(with A. Bostan & P. Lairez)

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Tt(f) :=⇣Ann f + @tQ(x, t)h@

x

, @ti| {z }int. by parts

⌘\ Q(x)h@

x

i| {z }di↵. under

R.

Q(x)h@x

, @tiQ(x)[t, 1/den f]h@

x

, @ti

Multiple binomial sums (with A. Bostan & P. Lairez)

Def. Combination of binomial coefficients, geometric sequences, +, x, multiplication by scalars, affine changes of indices and ∑.

Thm. The generating series is the integral of a rational function that can be constructed automatically.

→ whence an efficient summation algorithm using the previous algo.

Ex. (Apéry) A(n) =nX

k=0

✓n

k

◆2✓n+ k

k

◆2

7!I

dt1 ^ dt2 ^ dt3t1t2t3(1� t1t2 � t1t2t3)� (1+ t1)(1+ t2)(1+ t3)z

7! LDE 7! (n+ 1)3A(n) + (· · · )A(n+ 1) + (n+ 2)2A(n+ 2) = 0.

More at Pierre’s PhD defense on Nov. 12.

V. Numerical evaluation via the Taylor series

From large integers to precise numerical values

Numerical evaluation of solutions of LDEs

1. linear recurrence in N for the first sum (easy); 2. tight bounds on the tail (e.g., work with M. Mezzarobba); 3. no numerical roundoff errors.

f solution of a LDE with coeffs in Q(x) (our data-structure!)

f(x) =NX

n=0

anxn

| {z }fast evaluation

+1X

n=N+1

anxn

| {z }good bounds

The technique used for fast evaluation of constants like

1

⇡=

12

C3/2

1X

n=0

(�1)n(6n)!(A+ nB)

(3n)!n!3C3n

with A=13591409, B=545140134,

C=640320.

Code available: NumGfun [Mezzarobba 2010]

Principle:

Binary Splitting for linear recurrences (70’s and 80’s)

• n! by divide-and-conquer: Cost: O(n log3n loglog n) using FFT

• linear recurrences of order 1 reduce to

• arbitrary order: same idea, same cost (matrix factorial):

n! = n⇥ · · ·⇥ dn/2e| {z }size O(n log n)

⇥bn/2c ⇥ · · ·⇥ 1| {z }size O(n log n)

p!(n) := (p(n)⇥ · · ·⇥ p(dn/2e))⇥ (p(bn/2c)⇥ · · ·⇥ p(1))

ex: satisfies a 2nd order rec, computed via

✓enen�1

◆=

1

n

✓n+ 1 �1n 0

◆

| {z }A(n)

✓en�1

en�2

◆=

1

n!A!(n)

✓10

◆.

en :=nX

k=0

1

k!

Analytic continuation

Ex: erf(π) with 15 digits: 0 ������!

200 terms3.1416 �����!

18 terms3.1415927 �����!

6 terms3.14159265358979

arctan(1+i)

Again: computation on integers. No roundoff errors.

Compute as new initial conditions and handle error propagation:

f(x), f 0(x), . . . , f(d�1)(x)

VI Chebyshev expansions

Taylor Chebyshev

From equations to operators(n↦n+1) ↔ S mult by n ↔ n

composition ↔ product Sn=(n+1)S

Taylor morphism: D ↦ (n+1)S; x ↦ S-1

produces linear recurrence from LDE

Ore (1933): general framework for these non-commutative polynomials. Main property: deg AB=deg A+deg B. Consequence 1: (non-commutative) Euclidean division Consequence 2: (non-commutative) Euclidean algorithm.

d/dx ↔ D mult by x ↔ x

composition ↔ product Dx=xD+1

erf: D

2 + 2xD 7! (n+ 1)S(n+ 1)S+ 2S

�1(n+ 1)S = (n+ 1)(n+ 2)S2 + 2n

Ore fractions (Q-1P with P&Q operators)

Application: extend Taylor morphism to Chebyshev expansions

Thm. (Ore 1933) Sums and products reduce to that form.

Taylor xn+1=x·xn, (xn)’=nxn-1

↔ X:=S-1, D:=(n+1)S

Prop. [with A. Benoit] If y is a solution of L(x,d/dx), then its Chebyshev coefficients annihilate the numerator of L(X,D).

erf: D

2 + 2xD 7! (2(S�1 � S)�1n)2 + 2

S+ S

�1

2

2(S�1 � S)�1n

= pol(n, S)�1(2(n+ 1)(n+ 4)S4 � 4(n+ 2)3S2 + 2n(n+ 3))

Efficient numerical use: arXiv:1407.2802

Chebyshev 2xTn(x)=Tn+1(x)+Tn-1(x),

2(1-x2)Tn’(x)=-nTn+1(x)+nTn-1(x) ↔ X:=(S+S-1)/2,

D:=(1-X2)-1n(S-S-1)/2=2(S-1-S)-1n.

Next steps: FastRelax (starting this Fall)

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Computer Algebra Formal proofs

ComputerArithmetic

y

00 + 2xy

0 = 0 + ini. cond.

5 teams, 4 years

double erf(double x) {…}

Conclusion

• Linear differential equations and recurrences are a great data-structure;

• Numerous algorithms have been developed in computer algebra;

• Efficient code is available; • More is true (creative telescoping, diagonals,…); • More to come for Mireille and her followers.

Bravo Mireille !